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(C-D) According to the modified Maxwell-Ampere law for a world with magnetic charges,the magnetic $emf$ is given by $\oint B \cdot dl = \mu_{0} \epsil

Vedclass · Accepted Answer

$\frac{v E l}{c^{2}}$ counter-clockwise

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(C-D) According to the modified Maxwell-Ampere law for a world with magnetic charges,the magnetic $emf$ is given by $\oint B \cdot dl = \mu_{0} \epsilon_{0} \frac{d\phi_{E}}{dt}$.The electric flux $\phi_{E}$ through the circuit is $\phi_{E} = E \cdot A = E \cdot (l \cdot x)$,where $x$ is the position of the rod.The rate of change of electric flux is $\frac{d\phi_{E}}{dt} = E \cdot l \cdot \frac{dx}{dt} = E \cdot l \cdot v$.Substituting this into the $emf$ equation,we get $emf = \mu_{0} \epsilon_{0} E l v$.Since $c^{2} = \frac{1}{\mu_{0} \epsilon_{0}}$,we have $emf = \frac{E v l}{c^{2}}$.The direction of the induced magnetic current depends on the direction of the electric field $E$. If $E$ is directed outward,the current is clockwise; if $E$ is directed inward,the current is counter-clockwise. Thus,both options $(c)$ and $(d)$ are physically possible depending on the orientation of $E$.

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